\newproblem{lay:6_8_6}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.8.6}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let's define the inner product in the set of all continuous functions within the range $[0,2\pi]$ as
	\begin{center}
		$\left<f(t),g(t)\right>=\int\limits_0^{2\pi}{f(t)g(t)dt}$
	\end{center}
	Show that the functions $\sin(mt)$ and $\cos(nt)$ are orthogonal for all positive integers $m$ and $n$.
}{
   % Solution
	Let us solve first the indefinite integral
	\begin{center}
		$\int{\sin(mt)\cos(nt)dt}=-\frac{n\sin(mx)\sin(nx)+m\cos(mx)\cos(nx)}{m^2-n^2}$
	\end{center}
	Let us compute now the inner product
	\begin{center}
		$\begin{array}{rcl}\left<\sin(mt),\cos(nt)\right>&=&\int\limits_0^{2\pi}{\sin(mt)\cos(nt)dt}\\
		   &=&\left.-\frac{n\sin(mx)\sin(nx)+m\cos(mx)\cos(nx)}{m^2-n^2}\right|_0^{2\pi}\\
		   &=&-\frac{n\sin(2\pi m)\sin(2\pi n)+m\cos(2\pi m)\cos(2\pi n)}{m^2-n^2}-
			    \left(-\frac{n\sin(0)\sin(0)+m\cos(0)\cos(0)}{m^2-n^2} \right)\\
		   &=&-\frac{m}{m^2-n^2}-\left(-\frac{m}{m^2-n^2} \right)\\
			 &=&0
		\end{array}$
	\end{center}
	So the two functions $\sin(mt)$ and $\cos(nt)$ are orthogonal.
}
\useproblem{lay:6_8_6}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

